I sometimes think about the arrogant dwarf starsmiths in the question. How to blow up a star by accident?

I pose the question in the voice of the dwarf, probably because I am hungry for down votes.

A dead world, eons frozen, circling the dark of its dead sun. Long ago this world bloomed with life. But its sun was old. Even as the star withered in dying and its fires dimmed, its child moved away, as if fearful of the dying. Then the child died too. At the equator, last refuge in a freezing world, this world’s creatures lay yet, curled waiting for a summer that would never come. The strange snows cover their forms – first water, then nitrogen, then last the oxygen, falling from the cold sky. Thus the world has lain, for a thousand million years.

Could this world live again? Its own goddess is gone, but the Goddess of our home world is free with her favors. Just as a cave opened to the light will spring to the green, so too the Goddess could take this world and make it her own – if we could give her a light and heat to do it.

We do not have the craft to build a star but we have mastery of fission. We can build a moonlet of the heavy elements that sustain the fission. The fires of fission cannot be allowed to burn too hot - to stoke the fire, we will lard it with carbon and quench the force that would otherwise overwhelm it. Controlled, a great stone (50,000 m 3 ) of the ninety-second element (uranium) can burn with the power of a star for a million years. source

The power is there. Now how to make this a light suitable for a Goddess? The glow of a fusion star is its heat but this moon cannot be as hot as a star or it will burn too fast. Each element has its color. Can we add these elements such that when heated they glow with the reds and greens of a star, but without the heat that would consume it before the world is warmed?

In sum: A life-giving star emits much energy in the visible frequencies.

sun frequencies


A fission reaction hot enough to glow like a star will runaway and explode. Can one add various elements to the surface of a fissioning satellite of uranium such that the glow of the respective elements can duplicate the frequencies (and energy output) of a star?

  • 1
    $\begingroup$ It depends on how good an imitation of a black body radiation curve you want. Uranium melts at 1400 K; this is far from the 5900 K color temperature of the Sun. You may coat the hot metal globe with a suitable material which fluoresces with an anti-Stokes shift, thus converting some of the infrared photons into visible photons. For example, common LED lamps emit a spectrum very unlike a black body radiator, and yet our eyes find their light good enough. But if you want 100% fidelity then sorry, you must use a 5900 K black body. $\endgroup$
    – AlexP
    Commented Feb 18, 2018 at 22:50
  • $\begingroup$ I worry about photosynthesis, which requires the right wavelengths. If @AlexP is correct then you seem to have a problem, because you need to be close to 100% fidelity or you will have a wonderfully illuminated world that won't grow but a handful of plants. $\endgroup$
    – JBH
    Commented Feb 18, 2018 at 23:38
  • $\begingroup$ Are you essentially asking if this satellite can create a spectrum very close to that of a star? $\endgroup$
    – HDE 226868
    Commented Feb 19, 2018 at 0:01
  • 1
    $\begingroup$ Sorry to starve you, but I upvoted. $\endgroup$
    – kingledion
    Commented Feb 19, 2018 at 0:02
  • $\begingroup$ @HDE 226868 - that is exactly what I am asking but I care only about the visible spectrum. X-rays are of no use to the Goddess. I know it cannot be via blackbody radiation. I hope to duplicate the visual spectrum of a star using the emission spectra of various elements. $\endgroup$
    – Willk
    Commented Feb 19, 2018 at 0:50

2 Answers 2


Edited to object to the question

I guess I don't speak dwarf, so I didn't pick up the fact that you said that you can't have the sun that hot or fission will go too fast. In fact that is...backwards!

Lets say you have a solid mass of Uranium. This Uranium has two components, U-235 and U-238. U-235 is fissile, U-238 is not. So unless you have something like ~5% U-235 (which is called enriched Uranium), you will never start a fission chain reaction. Well...there are exceptions, but we will get to them.

In any case, lets say that you have an enriched uranium ball that is well above critical mass, being moon sized and all. So it is super-critical, and it starts producing energy. What stops it? First off, lets be clear that the particles ejected from a fission reaction have more than enough kinetic energy to escape the solar system. So if you do have a near-compelete chain reaction, your moon sized Uranium block will be turned into an expanding ball of plasma within hours. So your reaction has to be controlled in some way to not proceed as a chain reaction.

Second off, U-235 does not undergo fission from 'fast' neutrons. The neutrons spit out from a fission reaction will have some MeV of energy; if the hit a U-235 at that energy, they will simply bounce off, there will be no fission. They have to be going slow enough to react with the U-235 nucleus; they need to 'thermalize' down few orders of magnitude of energy. This is what would eventually stop the chain reaction; as the material gets blown apart, there is nothing to thermalize the escaping neutrons and the chain reaction will stop.

To lose this energy, neutrons have to bounce around like a pinball off of other uranium atoms (since there isn't anything else to react with in a moon sized block of uranium). Now, U-238 has a resonance absorption region, where at certain specific energies, a neutron that hit is will be captured. Going from MeV to keV will take many 'bounces' from a neutron, so it has a chance to get absorbed by U-238 on each bounce, if its energy is just right.

There is an effect called doppler broadening at high temperatures, where the breadth of the emission and absorption spectrum of an atom will increase when the atom has a high kinetic energy. So, as U-238 increases in temperature, it will increase in its ability to absorb neutrons. What his means in practice is that if your reaction is slow enough in the first place to not be a bomb, it will soon be shut down as the temperature increases.

So you put your problem exactly backwards; if your temperature gets high enough, you won't even have a star any more, just a warm rock.

Conclusion to objection

Your described mechanism for making the 'fission star' does not work as is; and your statement that increased temperature will make it burn too fast doesn't work either. Therefore, I am answering the question from the assumption that it works, somehow. If it works, somehow, then this is the best way to get a sun-like absorption.

Last note, I think there is a way to make a self-moderating block of Uranium, but that way is turning it into a U-238 fast-fission breeder reactor, where temperature regulation comes from setting the rate at which U-238 is bred into fissile Pu-239. But that is a subject for another question.

Back to generating spectra...


No need to fuss with special elements, the emission spectrum of a hot object is mostly based on its temperature. Here is the emission band from the sun.

enter image description here

The black line is the 5250 C blackbody radiation curve. This is the spectrum that anything of that temperature will emit.

You may be concerned about the spectral emission/abosorbtion lines for various elements, and how that would affect the light. Well, hydrogen has absorption lines at 410, 434, 486, and 656 nm. Those dips are barely visible on the graph. Sure there is a slight reduction at those specific frequencies, but not enough to make much of difference.

Uranium has absorption lines all over the place, but since Hydrogen's absorption didn't matter much, Uranium's won't either.

enter image description here


All you have to do is regulate temperature at the appropriate level, and you will have a sun-like light source. Now, how you will get your fusion star to stay at the correct temperature is non-trivial, but that sounds like a great follow-up question.

  • $\begingroup$ I thought WillK stated that he didn't want to have the object be the same temperature as a star exhibiting the desired spectrum ("his moon cannot be as hot as a star or it will burn too fast"). $\endgroup$
    – HDE 226868
    Commented Feb 19, 2018 at 0:17
  • $\begingroup$ @HDE226868 Thats the problem with writing your question in the voice of a dwarf... $\endgroup$
    – kingledion
    Commented Feb 19, 2018 at 1:09
  • $\begingroup$ Yes, dwarf voice has issues. My understanding is that rate of reaction in a regular reactor is controlled by carbon rods that absorb neutrons. If the reaction gets too hot that is fine for the reaction but the heat melts / destroys the stuff that is supposed to choke it down. I worried that the heat of a star would mean there was no way to prevent an uncontrolled chain reaction. Maybe not? I still hoped to use emission spectra to recreate blackbody radiation. $\endgroup$
    – Willk
    Commented Feb 19, 2018 at 20:58
  • 2
    $\begingroup$ @Willk The carbon rods actually provide moderation and cause the reaction to go faster. If the carbon rods are removed, the reaction stops because no enough neutrons are thermalized. So some reactors remove carbon rods to stop the reaction. To insert something stop or slow the reaction, you use boron, or if you have lots of money like the US Navy, hafnium. These are called 'poisons,' xenon is the most powerful one, though obviously not much use as a rod. $\endgroup$
    – kingledion
    Commented Feb 19, 2018 at 23:30

Option 1: Continuous spectrum

Your first choice would be to recreate the black body spectrum of the star. For a black body at a temperature $T$, the intensity $J(\nu)$ is $$J(\nu)=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/kT}-1}$$ where $h$, $c$ and $k$ are constants. For low-energy photons, $J(\nu)\approx\frac{2kT}{c^2}\nu^2$. For high-energy photons, $J(\nu)\approx\nu^3e^{-h\nu/kT}$. In all regimes, if you want to have a certain intensity at a certain frequency, you need a specific temperature. There's no way around it.

This means that if you want to duplicate a black body spectrum, you need to recreate the object's temperature. To mimic the Sun's spectrum, you object would have to have a surface temperature of about 5800 K. You've already stated, though, that you can't have that. Therefore, this option is off the table.

Option 2: Discrete spectrum with line broadening

Your second choice - the one you proposed - would be to take a bunch of atoms in different excited states and produce emission lines. Now, thanks to miscellaneous interactions and collisions, normally a spectrum becomes essentially continuous. The result is a spectrum that appears continuous.

Another way that a discrete spectrum can become more continuous is through line broadening. This can happen through a number of mechanisms.

  1. Natural broadening. The Heisenberg uncertainty principle means that there is uncertainty in the "lifetime" of an energy state - the amount of time an electron will spend in that state. The energy-time version of the uncertainty principle is $$\Delta t\Delta E>\hbar/2$$ meaning that a smaller uncertainty in time implies a larger uncertainty in energy. This leads to a wider range of emission frequencies - an effect called natural broadening.

    Unfortunately, natural broadening isn't effective. For instance, the natural broadening of the Lyman $\alpha$ hydrogen line is $$\frac{\Delta\lambda}{\lambda}\sim2\times10^{-8}$$ In other words, it's smeared out by only a tenth of a billionth of its original wavelength.

  2. Doppler broadening. Electrons and atoms move around because they have non-zero temperatures, and thus non-zero kinetic energy. Therefore, the photons they emit are subject to the Doppler shift. It turns out that Doppler broadening is much more effective than natural broadening; we get $$\frac{\Delta\lambda}{\lambda}\approx3\times10^{-7}\left(\frac{T}{1\text{ K}}\right)$$ for a hydrogen atom of temperature $T$. At 1000 K, for instance, we see a broadening of $\sim10^{-5}$. This is, of course, also disappointingly small. At visible wavelengths, you'd see broadening of perhaps one-thousandth of a nanometer or so.

Other types of broadening - pressure broadening through collisions, or Zeeman broadening through magnetic fields - give you similarly paltry results. Therefore, I don't think that you could easily get a continuous spectrum from adding some discrete components.


Neither of these options is particularly appealing. In short, it doesn't seem feasible to recreate a star's spectrum without raising the temperature high enough to reach temperatures akin to the surface of the Sun.


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